Business. The profit, in thousands of dollars, for a product is p(X)=-x^2+140x-4700 where x is the production level in hundreds of units. Find the production level that maximizes the profit. Find the maximum profit.

The maximum profit of $ _____ is achieved when ______ units are produced.

the max is on the axis of symmetry

... x = -b / 2a = -140 / (2 * -1)

plug the max production into p(X) to find the max profit

To find the production level that maximizes the profit, we need to determine the value of x that corresponds to the maximum value of the profit function p(x).

The profit function is given as p(x) = -x^2 + 140x - 4700.

To find the maximum value of p(x), we can use calculus. We need to take the derivative of p(x) with respect to x and set it equal to 0 to find the critical points. Then, we can use the second derivative test to determine which critical points correspond to a maximum.

1. Take the derivative of p(x) with respect to x:
p'(x) = -2x + 140.

2. Set p'(x) equal to 0 and solve for x:
-2x + 140 = 0.
-2x = -140.
x = -140 / -2.
x = 70.

This gives us the critical point x = 70.

3. Now, let's find the second derivative of p(x) to apply the second derivative test:
p''(x) = -2.

Since the second derivative is negative (-2), it means that the critical point x = 70 corresponds to a maximum.

4. Calculate the maximum profit:
To find the maximum profit, substitute the value of x = 70 into the profit function p(x):
p(70) = -(70)^2 + 140(70) - 4700.
p(70) = -4900 + 9800 - 4700.
p(70) = 0.

Therefore, the maximum profit is $0, and it is achieved when 70 units are produced.